Your instincts are right - your math is wrong. The volume is proportional to the integral of the square of the function. By the disk method, I get that the volume is
$$2 \pi \int_{3/2}^2 dx \, y^2=2 \pi \int_{3/2}^2 dx \, (4-x^2)$$
This evaluates to $2 \pi (11/24)$.
It's just the volume of the first one, because this is entirely below the second.
This is "obvious" if you visualise climbing the "ridgeline" from the origin to the second apex: this ridgeline passes through the first apex, because that is in the same direction from the origin but at half the distance. After you pass the first apex, the second pyramid keeps increasing but the first starts decreasing. So the faces of the two pyramids which lie along the $x$ axis coincide, as do those which lie along the $y$ axis; but the other faces of the first pyramid lie below the corresponding faces of the second.
If this is difficult to visualise you can confirm it algebraically, though it is a fair bit of work. To give one example: consider $(x,y)$ in the triangle with vertices at $(30,0)$, $(20,20)$ and $(10,10)$. One side of this triangle is the line with equation $x+2y=30$, and in the region we have $x+2y\ge30$.
The face of the first pyramid which lies above the region is part of the plane containing $(30,0,0)$, $(30,30,0)$ and $(10,10,10)$: it has equation
$$x+2z=30\ .$$
The face of the second pyramid which lies above the region is part of the plane containing $(0,0,0)$, $(30,0,0)$ and $(20,20,20)$, with equation
$$y-z=0\ .$$
Therefore we have
$$x+2y\ge30\quad\Rightarrow\quad \frac{30-x}{2}\le y\quad\Rightarrow\quad z_1\le z_2\ .$$
Best Answer
First note that the two spheres have equations: $$ x^2+y^2+z^2=a^2 \quad \mbox{and}\quad x^2+(y-a)^2+z^2=a^2 $$ and intersect on the plane $y=a/2$.
So the volume of the intersection region is made of two identical cups that have height $h=a/2$ and radius of the basis circle $b=\frac{\sqrt{3}}{2}a$ and we can use the formula for the volume of a cup to evaluate the volume as: $$ V=2 V_{cup}=2\cdot\frac{\pi a}{12}\left(\frac{9}{4}a^2+\frac{a^2}{4} \right)=2\cdot\frac{5\pi a^3}{24} $$
If you want to justify this result with an integral, you can do it with a simple integral noting that the single cup is a solid of revolution around the $y$ axis of the curve $z=\sqrt{a^2-y^2}$ in the $y-z$ plane. So the volume is: $$ V_{cup}=\int_{a/2}^a \pi z(y)^2 dy=\pi \int_{a/2}^a (a^2-y^2)dy= \frac{5 \pi}{24}a^3 $$
If you want necessarily a triple integral than the better is to note the cylindrical symmetry around the y axis and use cylindrical coordinates around it ( this means that we switch the name of $y$ and $z$ axis), so the volume of a single cup becomes: $$ V_{cup}=\int_{a/2}^a\int_0^{2\pi}\int_0^{\sqrt{a^2-y^2}}\rho \;d \rho\; d \varphi\; d y $$
that, with a bit of work, gives the same result.
The use of a triple integral in rectangular coordinates is possible but really masochistic.